Apostol Section 13.25 #13 - Conic Sections Question: Prove that a similarity transformation (replacing $x$ by $tx$ and $y$ by $ty$) carries an ellipse with center at the origin into another ellipse with the same eccentricity.
(The next questions go on to prove the same result (and converses) for hyperbolas.)
Please don't feel the need to read through all my attempts, I just wanted to show what I had tried and give some background on what has been covered in the book.
I think my question boils down to this: am I allowed to say "given some conic, align the coordinate axes so that the origin is at the center of the conic and the x-axis is aligned with the main axes of the conic" without any further justification?

My Attempt
This is easy to prove for an ellipse (or hyperbola) in standard Cartesian form:
$$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$$
Replacing $x$ and $y$ by $tx$ and $ty$ respectively, we have
$$\frac{(tx)^2}{a^2}+\frac{(ty)^2}{a^2(1-e^2)}=\frac{x^2}{\left(\frac a t\right)^2}+\frac{(y)^2}{\left(\frac a t\right)^2(1-e^2)}=1$$
Which which is another ellipse with a center at the origin with the same eccentricity. The problem is that the standard form only describes ellipses and hyperbolas with vertical directrices. Intuitively I understand that you could choose the coordinate axes so that any ellipse is described by this equation, but this has not been made explicit in the book.
I suppose one question worth asking, at this point, is if this is all Apostol wanted me to prove? Taking the original question literally, it seems my argument for choosing the coordinate axes is a bit of hand-waving to get at the other cases of ellipses.
EDIT: After discussing this with a professor at the university I attended, his interpretation was that Apostol's (high-level) intention was just to show that scaling the shape of an ellipse does not change the eccentricity, because eccentricity is a description of the shape and not the size. He thought, after reading the chapter, that I should consider only those curves in the ideal standard form for all questions such as this, and if necessary remark (although it is yet unproven in the book) that any conic is congruent to one in standard form. In this case, my original solution is adequate.
EDIT #2: It now seems clear that Apostol's intention was to consider only the standard cases, as the next question is Use the Cartesian equation which represents all conics of eccentricity $e$ and center at the origin to prove that these conics are integral curves of the differential equation $y'=(e^2-1)x/y$. Indeed, it seems that the integral curves of that differential equation are given by $\frac{x^2}{C}+\frac{y^2}{C(1-e^2)}=1$, which are really only the central conics at the origin with horizontal directrices, not all conics at the origin. Since, in that question, he does not make a distinction, I think he is just being atypically loose with his terminology.
 A: The general Equation of a ellipse whose axes are parallel to $X-Y$ axes is
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ and the eccentricity is given by $e=\sqrt{1-\dfrac{b^2}{a^2}}$ (I am assuming $a>b$).
If a similarity transformation occurs i.e, $(x,y)\to(tx,ty)$, then new equation of ellipse is $$\dfrac{(tx)^2}{a^2}+\dfrac{(ty)^2}{b^2}=1\ or \dfrac{x^2}{(a/t)^2}+\dfrac{y^2}{(b/t)^2}=1$$
So the eccentrity of the new ellipse $e'=\sqrt{1-\dfrac{(b/t)^2}{(a/t)^2}}=e$ 
In case Hyperbola just replace $b^2$ with $-b^2$ and remove the condition $(a>b)$. 
Hence proved.
A: A central conic with center at the origin takes the general form
$$ax^2+bxy+cy^2=1$$
(missing linear terms; why?). We are being asked to prove that
$$t^2(ax^2+bxy+cy^2)=1$$
also has the same eccentricity.
The rotation that will zero out the cross-term will be the same for both, a fair bit of algebra shows that for the rotation
$$\begin{align*}x&=p\cos\,\varphi-q\sin\,\varphi\\y&=p\sin\,\varphi+q\cos\,\varphi\end{align*}$$
taking the $\varphi$ given by
$$\tan\;\varphi=\frac{b}{a-c+\sqrt{b^2+(a-c)^2}}$$
does the trick.
Applying that transformation yields the conic
$$\frac{a+c+\sqrt{(a-c)^2+b^2}}{2}p^2+\frac{a+c-\sqrt{(a-c)^2+b^2}}{2}q^2=1$$
in the "unscaled" case, and with the pendant factor $t^2$ in the scaled case, which now looks a lot like the standard form you're accustomed to...
A: 
my question boils down to this: am I allowed to say "given some conic, align the coordinate axes so that the origin is at the center of the conic and the x-axis is aligned with the main axes of the conic" without any further justification?

If eccentricity is invariant under the operation of "align the coordinate axes so that ..." then calculations involving the eccentricity will not be affected by the choice of a conveniently aligned coordinate system.  The invariance can be seen by defining eccentricity in terms that are invariant and do not refer to position in a coordinate system.  For that it is enough to define major and minor axis invariantly.
The major axis is the longest chord of the ellipse, and the minor axis is the longest chord perpendicular to the major axis.  The lengths of these axes are invariant under rigid motion or reflection of the plane.  The ratio of major to minor axes is also invariant under similarity transformations such as $(x,y) \to (tx,ty)$. The eccentricity can be calculated from this ratio, so that it is both scale-invariant (because the ratio is) and isometry-invariant (because it is a function of the axes). 
